Exploring Holomorphic Functions in Complex Analysis

[Firepanda]

I'm not too sure how to show this. Perhaps if I show d(e^z)/dz = e^z then does this conclude that e^z is holomorphic on all of C?

 

[Dick]

Why does d/dz(e^z)=e^z show e^z is holomorphic? Or at least, why are you saying you aren't sure this shows it's holomorphic?

 

[Firepanda]

I suppose I should be using the cauchy riemann equations then?

I was told not to use limits here so I didn't want to use the base definition of holomorphic-ness.

 

[Firepanda]

Why does d/dz(e^z)=e^z show e^z is holomorphic? Or at least, why are you saying you aren't sure this shows it's holomorphic?

Because in my notes he mentioned the power series

 

[Dick]

I suppose I should be using the cauchy riemann equations then?

I was told not to use limits here so I didn't want to use the base definition of holomorphic-ness.

Sure, use Cauchy-Riemann. That's easy enough. Then you'd want to show e^(-z) is also holomorphic. Do you know the sum of holomorphic functions is also holomorphic? If not then you could just directly show cosh(z) is holomorphic using CR.

 

[Firepanda]

i tried to use C-R, but I'm unable to split it up into

U(x,y) or V(x,y)

using z=x+iy into cosh(z)

 

[Dick]

i tried to use C-R, but I'm unable to split it up into

U(x,y) or V(x,y)

using z=x+iy into cosh(z)

How about e^z? Can you split that up?

 

[Firepanda]

How about e^z. Can you split that up?

e^xe^iy

= e^x(cosy + i.siny)

=e^xcosy + i.e^xsiny

The C-R satisfy this and so it is holomorphic on all of C? so this would immediatly imply cosh(z) is entire?

And I would check if it were true for e^-z as well.

 

[Dick]

e^xe^iy

= e^x(cosy + i.siny)

=e^xcosy + i.e^xsiny

The C-R satisfy this and so it is holomorphic on all of C? so this would immediatly imply cosh(z) is entire?

And I would check if it were true for e^-z as well.

Sure, check e^(-z) as well. It's almost the same thing. I'm a little puzzled why you can split e^(z) and e^(-z) up and not cosh(z). You already said cosh(z)=(e^z+e^(-z))/2.

 

[Firepanda]

Sure, check e^(-z) as well. It's almost the same thing. I'm a little puzzled why you can split e^(z) and e^(-z) up and not cosh(z). You already said cosh(z)=(e^z+e^(-z))/2.

ah ye that would make more sense if i did it directly, thanks!